I was chatting to Craig “Cap. Astro” Pastro yesterday evening over a beer, and he mentioned an interesting question: if you take the monoidal centre of a braided monoidal category, do you get back the thing you started with? We figured out the answer at breakfast this morning, and I’ll explain it in a bit. But first maybe I should explain what the question means.

The centre of a monoid M is defined to be the submonoid consisting of those elements that commute with everything in M. The centre of a monoid is always abelian (for obvious reasons!) and if M was abelian to start with, then the centre is just the whole of M.

If you categorify this, you get the centre construction on a monoidal category. The centre of the monoidal category C is the braided monoidal category in which an object consists of an object A of C together with a natural isomorphism

A⊗– → –⊗A

satisfying the usual braiding axioms; and the rest of the structure is defined in the only reasonable way. (If you’re not satisfied with that description, read on…)

Later today, Eugenia Cheng told me a nice way to think about the centre construction and its generalisation. The centre of a monoid can be described as follows: take your monoid M and think of it as a one-object category Σ(M); the centre is precisely the monoid of natural transformations from the identity functor on Σ(M) to itself. When you look at it like this, it’s easy to categorify: take your monoidal category C, and think of it as a one-object bicategory Σ(C). Then its centre is the monoidal category whose objects are pseudo-natural transformations from the identity pseudofunctor on Σ(C) to itself, and whose arrows are modifications between them. (The tensor product is given by composition, of course.) If you work out what this means, you get back the concrete description I sketched above.

Back to the question: we know that the centre of an abelian monoid M is isomorphic to M, so it’s natural to wonder whether the centre of (the monoidal category underlying) a braided monoidal category C is always braided monoidal equivalent to C. But if you think about what that would mean, it starts to sound pretty unlikely. For one thing, it would mean that braidings are essentially unique where they exist, and that doesn’t sound right.

Indeed it isn’t right, and here’s our counterexample. The idea is to construct a monoidal category that has two braidings, one of which is a symmetry and the other of which is not. Since braided monoidal equivalence preserves symmetry, these braidings must be inequivalent; therefore the centre must differ from at least one of them. Let B be the free symmetric strict monoidal category on one generator: so an object is just a natural number, and a morphism m → n is a bijection from the m-element set to the n-element-set. (If m ≠ n then there is no such morphism of course, so in fact the only morphisms are endomorphisms. In other words, B is the categorical coproduct of all the symmetric groups.) Let Z be the abelian group of integers under addition, regarded as a category with one object. You can think of Z as a symmetric monoidal category, where the tensor of two arrows (integers) is again their sum. It follows therefore that the product B×Z is a symmetric monoidal category, with symmetry σ, say. The objects of B×Z are just natural numbers again, and an arrow is a bijection together with an integer. When you compose or tensor two morphisms, you just add the integers together. Now we can define another braiding β where, for every pair of objects m, n, the arrow βm,n has the same bijection as σm,n and the associated integer is the product mn. If you check the axioms, you find that this is indeed a braiding, essentially because the product of integers distributes over their sum. And it can’t be symmetric, because the integer associated with the inverse of a β map is negative. So we’re done!